作者angie0716 (大宝与困宝)
看板GMAT
标题[计量] OG的168与175题
时间Tue Jul 22 16:35:35 2008
题目是这样的
168.An airline passenger is planning a trip that involves three connecting
flights that leaves from Airports A, B, and C,respectively. The first flight
leaves Airport A every hour, beginning at 8:00 a.m., and arrives at Airport
B 2 and 1/2 hours later. The second flight leaves Airport B every 20 minutes,
beginning at 8:00 a.m., and arrives at Airport C 1 and 1/6 hours later. The
third flight leaves Airport C every hour, beginning at 8:45 a.m. What is the
least total amount of time tha passenger must spend between flights if all
flights keep to theur schedules? Ans:1 hr 15 minutes
这题其实题目有点看不太懂..外加解答看的雾煞煞..
另一题175.
A square countertop has a square tile inlay in the center, leaving an untiles
strip of uniform width around the tile. If the ratio of the tiled area to the
untiled area is 25 to 39, which of the following could be the width, in inches
, of the strip?
Ans:1.3.4
这题也是题目看不太懂@@....麻烦大家了
谢谢:)
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1F：推 seawolfssn:168. A->B->C 三段航程联票 就是A出发到B B出发到C 07/22 16:53
2F：→ seawolfssn:请问旅客所花费的"候机"时间? 07/22 16:54
3F：→ seawolfssn:航站A 每小时一班飞机飞B 07/22 16:55
4F：→ seawolfssn:航站B 每20分钟一班飞机飞C 07/22 16:56
5F：→ seawolfssn:假设蝙蝠侠 在航站A搭乘早上8:00AM经过2 1/2小时到航B 07/22 16:59
6F：→ seawolfssn:他到达航站B 时间为10:30 所以蝙蝠侠比须等待 10分钟 07/22 17:00
7F：→ seawolfssn:搭乘10:40的飞机到航站C 07/22 17:00
8F：→ seawolfssn:蝙蝠侠花了1 1/6到达航站C 到达航站C时间为11:50 07/22 17:02
9F：→ seawolfssn: 小时 07/22 17:02
10F：→ seawolfssn:然後蝙蝠侠必须等12:45离开C航厦的班机 07/22 17:04
11F：→ seawolfssn:因为C航站从 8:45起 每一小时发一班飞机 07/22 17:05
12F：→ seawolfssn:所以蝙蝠侠必须等待 55分钟的候机时间~ 07/22 17:05
13F：→ seawolfssn:所以 10 + 55 = 65 = 1 hour 5 minutes 07/22 17:06
14F：→ seawolfssn:*补充* A->B->C 三段航程联票=A到B B到C C再到某地! 07/22 17:08